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Wave Equation

1-D Wave Equation

y(x,t)2yt2=c22yx2

Separate Variable Method

  • 前提

wave equation is on a closed interval [0,L]

assume that

y(x,t)=X(x)T(t)

then

Xn(x)=sin(nπxp)Tn(x)=ancos(nπcpt)+bnsin(nπcpt)

finally we have

y(x,t)=n=1yn(x,t)=n=1Xn(x) Tn(t)

zero initial velocity

yt(x,0)=0

let f(x) be an initial displacement function

f(x)=y(x,0)

then we can get

Tn(t)=ancos(nπcpt)

and finally,

y(x,t)=n=1ansin(nπxp)cosnπcpt
  • notice that p is a half of period ([p,p])

in which an is the sine Fourier series coefficient of f(x)=y(x,0)=ansin(nπxp)

an=2p0pf(x)sin(nπxp)dx

zero initial displacement

y(x,0)=0

let g(x) be the initial velocity function

g(x)=yt(x,0)

then we can get

Tn(t)=bnsin(nπcpt)

in which nπcpbn is the Fourier sine series coefficient of g(x)

nπcpbn=2p0pg(x)sin(nπxp)dx

finally,

y(x,t)=n=1yn(x,t)=n=1bnsin(nπxp)sin(nπctp)

nonzero initial velocity and displacement

in the situation, assume

y(x,t)=y1(x,t)+y2(x,t)

which y1(x,t) assume zero initial velocity, and that y2(x,t) assume zero initial displacement

and that is same as what we have discussed above.


with External Force

consider wave equation in this form

2yt2=2yx2+Ax

y(x,t) can not be separated.

so we let

y(x,t)=Y(x,t)+ϕ(x)

assume that Y(x,t) can be separated, which means we can get the answer with the method above.

代入 y(x,t), we can get

2Yt2=2Yt2+2ϕt2+Ax

therefore

ϕ(x)+Ax=0

Infinite Medium

for the wave equations which are define in (,)

as usual, denote the initial displacement as f(x) and the initial velocity as g(x)

first we have to know that the eigenvalue only exists when ω0

  • solve two ODE we can get
Xω(x)=aωcos(ωx)+bωsin(ωx)Tω(t)=Aωcos(ωct)+Bωsin(ωct)

then the most important thing is that

y(x,t)=0yω(x,t) dω=0Xω(x)Tω(t) dω

Semi-infinite Medium

for the wave equations which are define in [0,), thus

y(0,t)=0 for t>0

as usual, denote the initial displacement as f(x) and the initial velocity as g(x)

X(x)=acos(ωx)+bsin(ωx)T(t)=cωcos(ωct)+dωsin(ωct)

%% since X(0)=0, thus

X(x)=bsin(ωx)

%%


Finite V.S. Infinite Medium

Finite Infinite
Fourier sine series Fourier integral
nπp ω
X(x) sin(nπxp) acos(ωx)+bsin(ωx)
T(t) when v(0)=0 ancos(nπcpt) Aωcos(ωct)
T(t) when d(0)=0 bnsin(nπcpt) Bωsin(ωct)

Characteristics and d'Alembert's Solution

one special solution for these condition

utt=c2uxx<x<,t>0u(x,0)=f(x)<x<ut(x,0)=g(x)<x<

u(x,t)=12[f(xct)+f(x+ct)]+12cxctx+ctg(k) dk

Forced Wave Motion

for <x<,t>0utt=c2uxx+F(x,t)u(x,0)=f(x)ut(x,0)=g(x)
  • solution
u(x,t)=12[f(xct)+f(x+ct)]+12cxctx+ctg(x) dx+12cΔF(x,t) dx dt

Δ is characteristics triangle

xct=k1x+ct=k2

k1, k2 can be any real constants.


2-D Wave Equation

zt2=c2(2zx2+2zy2)